5! = 5 × 4 × 3 × 2 × 1 = 120. Why we multiply: each position reduces choices.

In How Many Ways Can 5 People Sit in a Row? — Permutation Basics

Question

In how many ways can 5 people sit in a row?

This is the foundational permutation question — every student meets it in Class 11, and it’s the building block for every harder arrangement problem in JEE and boards.

Solution — Step by Step

We have 5 seats in a row. The key insight: we’re not filling all seats at once — we fill them one by one, and each choice affects the next.

Start with the leftmost seat. Any of the 5 people can sit there, so we have 5 choices.

Once someone sits in Seat 1, they’re committed — they can’t be in Seat 2. So for the second seat, we only have 4 remaining people to choose from.

This is the core logic of permutations: each placement removes one option.

Seat 3: 3 people left → 3 choices

Seat 4: 2 people left → 2 choices

Seat 5: only 1 person remains → 1 choice

By the Fundamental Principle of Counting, when we make a series of independent decisions, we multiply the number of choices at each stage:

5 \times 4 \times 3 \times 2 \times 1 = 120

This product has a name: 5! (5 factorial). The general formula for arranging $n$ distinct objects in a row is $n!$ .

The answer is 120 ways.

Why This Works

The multiplication here isn’t arbitrary — it comes from the Fundamental Principle of Counting. If Task A can happen in $m$ ways and Task B can happen in $n$ ways after that, together they happen in $m \times n$ ways.

Filling a row of seats is exactly this: 5 sequential tasks (one per seat), each with a shrinking number of options. Multiply them and you get the total.

P = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For our problem: $5! = 120$ . Every distinct arrangement — say, A-B-C-D-E versus A-C-B-D-E — is counted exactly once.

Alternative Method

We can also use the permutation formula directly. Arranging 5 people in 5 seats is the same as selecting 5 people from 5 and arranging them:

^5P_5 = \frac{5!}{(5-5)!} = \frac{5!}{0!} = \frac{120}{1} = 120

We use $0! = 1$ by definition (there’s exactly one way to arrange zero objects — do nothing). This connects the seat-filling logic to the standard $^nP_r$ formula you’ll use for tougher problems like “how many ways to choose and arrange 3 people from 7.”

When all $n$ objects fill all $n$ positions, $^nP_n$ always equals $n!$ . No need to apply the full formula — just write $n!$ directly and save time in exams.

Common Mistake

Adding instead of multiplying. Some students write $5 + 4 + 3 + 2 + 1 = 15$ and think that’s the answer. This would be correct if you were choosing a seat for one person with 15 total options — but here we’re filling all seats for all people simultaneously. Separate sequential decisions must be multiplied, not added.

A quick sanity check: with just 2 people (A and B) in 2 seats, there are clearly 2 arrangements (AB and BA). Our formula gives $2! = 2$ ✓. With 3 people: ABC, ACB, BAC, BCA, CAB, CBA — that’s 6, and $3! = 6$ ✓. The multiplication logic holds.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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